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If a1,a2,.. ....... .are in GP with common ratio r such that (summation of k=1 to n) a(2k-1)=(summation of k=1 to n) a(2k+2)≠0.then possible values of r is.......

Pawan joshi , 6 Years ago
Grade 12th pass
anser 2 Answers
Susmita

Last Activity: 6 Years ago

Are you sure that it is not sum k=0 to n (2k+2)?
Whatever look at the solution.
2k-1=1 for k=1
        =3 for k=2
        =5 for k=3 and so on
2k+2=4 for k=1
        =6 for k=2
        =8 for k=3 and so on.
\sum_{1}^{n} a_{2k-1}=\sum_{1}^{n}a_{2k+2}
Or,a_{1}+a_{3}+a_{5}+...=a_{4}+a_{6}+a_{8}+...
Or,a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+...=-a{2}
So r=a2/a1=a3/a2=..=-1 and series sum is equal to -a2.
If it helps please approve.
 

Susmita

Last Activity: 6 Years ago

Now look series sum for infinite go series is
S=a1/(1-r)=a1/2 in our case.
But we also obtained that S=-a2
So a1/2=-a2
Or,a2=-a1/2
So r=a2/a1=-1/2
 

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